What are the vertices, foci, and asymptotes of the hyperbola with the equation 16x2–4y2=64 ?

Answer 1

The given equation of the hyperbola is in the standard form:

16x^2 - 4y^2 = 64

Dividing through by 64:

x^2/4 - y^2/16 = 1

This is a hyperbola with a horizontal transverse axis. The general equation for a hyperbola with a horizontal transverse axis is:

(x - h)^2/a^2 - (y - k)^2/b^2 = 1

Comparing this with the given equation, we have:
h = 0, k = 0, a^2 = 4 (so a = 2), b^2 = 16 (so b = 4)

Therefore, the center of the hyperbola is at (0, 0), the vertices are (±2, 0), the foci are (±√20, 0), and the asymptotes are given by the equations y = ±(b/a)x, which are y = ±2x.

So, the vertices are (±2, 0), the foci are (±√20, 0), and the asymptotes are y = ±2x.